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Physics 2D Lecture Slides Mar 10 Vivek Sharma UCSD Physics QM in 3 Dimensions • Learn to extend S. Eq and its solutions from “toy” examples in 1-Dimension (x) three orthogonal dimensions (rx,y,z) ˆ ˆ ˆjy kz r ix • Then transform the systems – Particle in 1D rigid box 3D rigid box – 1D Harmonic Oscillator 3D Harmonic Oscillator z • Keep an eye on the number of different integers needed to specify system 1 3 (corresponding to 3 available degrees of freedom x,y,z) y x Quantum Mechanics In 3D: Particle in 3D Box z Extension of a Particle In a Box with rigid walls 1D 3D Box with Rigid Walls (U=) in X,Y,Z dimensions z=L U(r)=0 for (0<x,y,z,<L) Ask same questions: • Location of particle in 3d Box • Momentum • Kinetic Energy, Total Energy • Expectation values in 3D y y=0 y=L x To find the Wavefunction and various expectation values, we must first set up the appropriate TDSE & TISE The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates Time Dependent Schrodinger Eqn: 2 2m 2 ( x, y , z , t ) U ( x, y , z ) ( x, t ) i ( x, y, z , t ) .....In 3D t 2 2 2 2 2 2 x y z 2 z 2 2 2 2 2 2 So [K ] 2 2 2 2m 2m x 2m y 2m z = [K x ] + [K y ] + [K z ] 2 2 y x so [ H ] ( x, t ) [ E ] ( x, t ) is still the Energy Conservation Eq Stationary states are those for which all probabilities are constant in time and are given by the solution of the TDSE in seperable form: ( x, y, z , t ) (r , t ) = (r)e -i t This statement is simply an extension of what we derived in case of 1D time-independent potential Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables 2 TISE in 3D: - 2 ( x, y, z ) U ( x, y, z ) ( x, y, z ) E ( x, y, z ) 2m x,y,z independent of each other , write ( x, y, z ) 1 ( x) 2 ( y) 3 ( z ) and substitute in the master TISE, after dividing thruout by = 1 ( x) 2 ( y ) 3 ( z ) and noting that U(r)=0 for (0<x,y,z,<L) 2 2 2 1 2 1 ( x) 1 2 2 ( y ) 1 2 3 ( z ) E Const 2 2 2 2m 1 ( x) x 2m 2 ( y ) y 2m 3 ( z ) z This can only be true if each term is constant for all x,y,z 2 2 2 3 ( z ) 2 1 ( x) 2 2 ( y ) E1 1 ( x) ; E2 2 ( y ) ; E3 3 ( z ) 2 2 2 2m x 2m y 2m z 2 With E1 E 2 E 3 E=Constant (Total Energy of 3D system) Each term looks like particle in 1D box (just a different dimension) So wavefunctions must be like 1 ( x) sin k1x , 2 ( y ) sin k2 y , 3 ( z ) sin k3 z Particle in 3D Rigid Box : Separation of Orthogonal Variables Wavefunctions are like 1 ( x) sin k1x , 2 ( y ) sin k2 y , 3 ( z ) sin k3 z Continuity Conditions for i and its first spatial derivatives ni ki L Leads to usual Quantization of Linear Momentum p= k .....in 3D px L n1 ; p y n2 ; pz n3 (n1 , n 2 , n 3 1, 2,3,..) L L Note: by usual Uncertainty Principle argument neither of n1 , n 2 , n 3 0! ( why ?) 1 2 2 2 2 2 2 2 2 Particle Energy E = K+U = K +0 = ( px p y pz ) (n1 n2 n3 ) 2 2m 2mL Energy is again quantized and brought to you by integers n1 , n 2 , n 3 (independent) and (r)=A sin k1x sin k2 y sin k3 z (A = Overall Normalization Constant) (r,t)= (r) e E -i t A [sin k1x sin k2 y sin k3 z ] e E -i t Particle in 3D Box :Wave function Normalization Condition (r,t)= (r) e E -i t * (r,t)= * (r) e A [sin k1x sin k2 y sin k3 z ] e E i t E -i t A [sin k1x sin k2 y sin k3 z ] e E i t * (r,t) (r,t)=A2 [sin 2 k1x sin 2 k2 y sin 2 k3 z ] Normalization Condition : 1 = P(r)dx dydz x,y,z L L L 1 A sin k1x dx sin k2 y dy sin k3z dz = A x=0 y=0 z=0 2 2 2 L L 2 2 3 2 2 A and ( r,t)= L L 2 2 3 2 2 E i t 2 L [sin k1x sin k2 y sin k3 z ] e Particle in 3D Box : Energy Spectrum & Degeneracy E n1 ,n 2 ,n 3 2 2 2 2mL ( n12 n22 n32 ); n i 1, 2, 3..., ni 0 Ground State Energy E111 3 2 2 2mL2 6 2 2 Next level 3 Excited states E 211 = E121 E112 2mL2 Different configurations of (r)= (x,y,z) have same energy degeneracy z z=L y=L y x=L x Degenerate States E 211 = E121 E112 Ground State 6 2 2 2mL2 E111 z z y y x x E211 E121 E112 Probability Density Functions for Particle in 3D Box Same Energy Degenerate States Cant tell by measuring energy if particle is in 211, 121, 112 quantum State • Degeneracy came from the threefold symmetry of a CUBICAL Box (Lx= Ly= Lz=L) • To Lift (remove) degeneracy change each dimension such that CUBICAL box Rectangular Box • (Lx Ly Lz) • Then n12 2 2 n22 2 2 n32 2 2 E 2 2 2 2 mL 2 mL 2 mL x y z Energy Source of Degeneracy: How to “Lift” Degeneracy The Hydrogen Atom In Its Full Quantum Mechanical Glory U (r ) r 1 r 1 x y z 2 2 2 More complicated form of U than box By example of particle in 3D box, need to use seperation of variables(x,y,z) to derive 3 independent differential. eqns. This approach will get very ugly since we have a "conjoined triplet" To simplify the situation, use appropriate variables Independent Cartesian (x,y,z) Inde. Spherical Polar (r, , ) 2 2 2 Instead of writing Laplacian 2 2 2 , write x y z 2 1 2 1 1 2 = 2 r 2 sin 2 2 r r r r sin r sin 2 2 TISE for (x,y,z)= (r, , ) becomes kZe2 U (r ) r 1 2 (r, , ) 1 (r, , ) r sin 2 2 r r r r sin 1 2 (r, , ) 2m + 2 (E-U(r)) (r, , ) =0 r 2 sin 2 2 !!!! fun!!! Spherical Polar Coordinate System Volume Element dV dV (r sin d )(rd )(dr ) = r 2 sin drd d Don’t Panic: Its simpler than you think ! 1 2 1 1 2 2m + 2 (E-U(r)) (r, , ) =0 r sin r 2 r r r 2 sin r 2 sin 2 2 Try to free up last term from all except This requires multiplying thruout by r 2 sin 2 2 2 2mr 2 sin 2 ke 2 sin + (E+ ) =0 r sin sin 2 r r 2 r For Seperation of Variables, Write (r, , )=R(r).( ). ( ) 2 Plug it into the TISE above & divide thruout by (r, , )=R(r).( ). ( ) Note that : ( r , , ) R(r) ( ). ( ) r r ( r , , ) ( ) R ( r ) ( ) when substituted in TISE ( r , , ) ( ) R ( r )( ) sin 2 2 R sin 1 2 2mr 2 sin 2 ke 2 + (E+ ) =0 r sin 2 R r r 2 r Rearrange by taking the term on RHS sin 2 2 R sin 2mr 2 sin 2 ke 2 1 2 (E+ ) =r sin + 2 R r r r 2 LHS is fn. of r, & RHS is fn of only , for equality to be true for all r, , LHS= constant = RHS = ml2 Now go break up LHS to seperate the r & terms..... sin 2 2 R sin 2mr 2 sin 2 ke 2 LHS: (E+ ) =ml2 r sin + 2 R r r r Divide Thruout by sin 2 and arrange all terms with r away from ml2 1 2 R 2mr 2 ke 2 1 )= 2 r 2 (E+ sin R r r r sin sin Same argument : LHS is fn of r, RHS is fn of , for them to be equal for all r, LHS = const = RHS = l (l 1) What do we have after shuffling! d 2 2 m 0...................(1) l 2 d ml2 1 d d sin l (l 1) 2 ( ) 0.....(2) sin d d sin 1 d 2 R 2mr 2 ke 2 l (l 1) )R(r ) 0....(3) r 2 (E+ 2 2 r dr r r r These 3 "simple" diff. eqn describe the physics of the Hydrogen atom. All we need to do now is guess the solutions of the diff. equations Each of them, clearly, has a different functional form